Counting Lines; Hydrogen-like Ions

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Hydrogen atoms in a tube are excited up to level four and fall back to the ground state. How many different spectral lines do we see? Four to three, three to two, two to one, so three lines, right? Hold on to that answer. Today we count properly, then see how helium plus and lithium two plus copy hydrogen. This comes up often in JEE, so let's begin.

0:30Many atoms: one line for every pair of levels

Here is the level diagram of hydrogen from the last lesson. Each horizontal line is a level, labelled by n, the principal quantum number, and each downward arrow is one jump, giving one spectral line; the arrows from the dashed top line, n equal to infinity, only mark the series limits. From level four, the electron can jump to three, to two, or straight to one; the little arrow on screen means such a jump. Okay? But wait, think about an atom that dropped from four to three. Is that atom finished, or can it give more lines? It is not finished, because it is still excited. From level three it can fall to two or to one, and from level two it falls to one. That gives three more lines. Right? So the total is three plus two plus one, six lines, not three! With so many atoms, every possible jump happens somewhere. Now, each line joins one pair of levels. From n levels, how many pairs can you make? Yes, exactly, n times n minus one, over two, with n minus one in brackets. So atoms excited to level n emit that many different lines in all. Check with four: four times three is twelve, and twelve over two gives six. Clear? Now, lines between an upper level n two and a lower level n one. Here n two has a small two below the n, and n one has a small one below it. There are n two minus n one, plus one levels in play, so we get n two minus n one, times n two minus n one plus one, over two lines. With me?

2:08One atom: only one path down

Now, a different question. Not a tube full of atoms, just one single hydrogen atom in level four. Can this one atom give all six lines? No, it cannot, because one atom takes only one path down. It gives the most photons by dropping one level at a time: four to three, three to two, two to one. So one atom in level n emits at most n minus one photons. Okay? Now the least number. The atom can jump straight down to n equal to one in a single step, so at least one photon. Got it? So for level four: at most three photons, and at least one. See, that is where the puzzle's answer of three was hiding: three is the most that one atom can give, while many atoms give six. So always check: one atom, or many? Clear?

2:59Hydrogen-like ions: everything scales with Z²

Now the second idea. Helium plus, H e with a small plus up top, and lithium two plus, L i with a small two plus up top, have just one electron each, like hydrogen. These are hydrogen-like ions, and Z, the atomic number, is two for helium and three for lithium. With me? From the last lesson, the photon energy h nu, with h, Planck's constant, and nu, the Greek letter for frequency, is thirteen point six times Z squared, times the bracket, one over n one squared minus one over n two squared, in electron volts, the e V on screen. The small two up top means squared. Okay? Now keep the same two levels, but change Z from one to two. What happens to the photon energy, and to the wavelength? Yes, exactly, the energy becomes Z squared times bigger: four times for helium plus, nine times for lithium two plus. And the wavelength goes as one over the energy, so every wavelength is divided by Z squared. The cross is the times sign, and the line with two dots is the division sign. Right? Now look at the diagram, hydrogen on the left and helium plus on the right. Level two of helium plus sits at minus thirteen point six electron volts, level with the ground state of hydrogen; see the arrow marked equal. Z squared and n squared are both four, so Z squared over n squared is one, as for the ground state of hydrogen. So the helium plus line from two times n two, down to two times n one, matches the hydrogen line from n two to n one, capital H for hydrogen. Clear? For example, helium plus from four to two matches hydrogen from two to one, the first Lyman line, ending on level one. Its wavelength is one hundred and twenty one point six nanometres, the n m on screen. Got it?

4:48Example 1: lines from n = 5 and how many are visible

Now, Example one. Hydrogen atoms are excited to the level n equal to five. How many different lines can be emitted, and how many lie in the visible region? Pause the video and try it first. Done? Okay, let's check. Many atoms, so we use n times n minus one, over two. Five times four is twenty, and twenty over two gives ten lines. Okay? Now the visible region. Think back to the last lesson: which series gives the visible lines? Very good, it is the Balmer series, and its lines all end on n equal to two. Right? So we need the atoms that land on level two, those that start from three, four or five, so three visible lines, out of the ten. With me?

5:38Example 2: find Z from 91.8 eV, then the series limit

Now a harder one. In a hydrogen-like ion, the photon of the transition from n equal to two, to one, has energy ninety one point eight electron volts. Find Z, and the wavelength of the series limit of its Balmer-type series, ending on n equal to two. Pause the video and try it first. Done? Okay, let's check. Many students write thirteen point six times Z squared equals ninety one point eight. Then Z squared is ninety one point eight over thirteen point six, which gives six point seven five. Is that fine? That is wrong, so be careful here, because Z must be a whole number. Thirteen point six times Z squared alone is the jump from level one all the way out. For the jump from two to one, multiply by the bracket, one minus one quarter, which is three quarters. That gives ten point two times Z squared, equal to ninety one point eight. Clear? So Z squared is ninety one point eight over ten point two, which is nine, and Z is three. That is lithium two plus. Okay? Now the series limit, the line with the shortest wavelength. It ends on n equal to two, but which upper level does it start from? Yes, from infinity, the sideways eight on screen, so one over n two squared becomes zero. The energy is thirteen point six times nine, over four. That is one hundred and twenty two point four over four, which gives thirty point six electron volts. Got it? Now the wavelength, lambda, the Greek letter. From the last lesson, lambda in nanometres is one thousand two hundred and forty over the energy in electron volts. So one thousand two hundred and forty over thirty point six gives forty point five nanometres. With me?

7:32Quick revision: many atoms n(n-1)/2, one atom n-1, Z squared

So, quick revision. Many atoms up to level n give n times n minus one, over two, lines. One atom in level n gives at most n minus one photons, and at least one. And for hydrogen-like ions, every energy is multiplied by Z squared, and every wavelength divided by Z squared. Now go and try the practice problems on this page. All the best!